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Chapter 9 Linear Momentum, Systems of Particles, and Collisions Linear momentum (Ch. 4) • Linear momentum (or, simply momentum) point-like object (particle) is p of a p mv • SI unit of linear momentum is kg*m/s • Momentum is a vector, its direction coincides with the direction of velocity Newton’s Second Law revisited (Ch. 4) • Originally, Newton formulated his Second Law in a more general form Fnet dp dt • The rate of change of the momentum of an object is equal to the net force acting on the object • For a constant mass Fnet dp d (mv ) dv m ma dt dt dt Center of mass • In a certain reference frame we consider a system of particles, each of which can be described by a mass and a position vector • For this system we can define a center of mass: rCM mi ri i m i i mi ri i M Center of mass of two particles • A system consists of two particles on the x axis • Then the center of mass is rCM mi ri i m i mi ri i M i xCM m1 x1 m2 x2 m1 m2 yCM m1 0 m2 0 m1 m2 0 Center of mass of a rigid body • For a system of individual particles we have rCM mi ri i m i i • For a rigid body (continuous assembly of matter) with volume V and density ρ(V) we generalize a definition of a center of mass: rCM volume r dV volume dV r dm M Chapter 9 Problem 41 Find the center of mass of the uniform, solid cone of height h, base radius R, and constant density shown in the figure. (Hint: Integrate over disk-shaped mass elements of thickness dy, as shown in the figure.) Newton’s Second Law for a system of particles • For a system of particles, the center of mass is rCM mi ri i m i • Then i MrCM mi ri MaCM i d mi ri i 2 dt 2 d ri mi 2 dt i 2 mi ri i M 2 d rCM d ( MrCM ) M 2 dt dt 2 2 F mi ai i i i Newton’s Second Law for a system of particles • From the previous slide: MaCM Fi i • Here Fi is a resultant force on particle i • According to the Newton’s Third Law, the forces that particles of the system exert on each other (internal forces) should cancel: MaCM Fnet • Here Fnet is the net force of all external forces that act on the system (assuming the mass of the system does not change) Newton’s Second Law for a system of particles MaCM Fnet Linear momentum for a system of particles • We define a total momentum of a system as: P pi mi vi i i • Using the definition of the center of mass dri P mi vi mi dt i i d mi ri i dt drCM M dt MvCM • The linear momentum of a system of particles is equal to the product of the total mass of the system m r m r and the velocity of the center of mass ii ii rCM i m i i i M Linear momentum for a system of particles • Total momentum of a system: P MvCM • Taking a time derivative dvCM dP MaCM Fnet M dt dt • Alternative form of the Newton’s Second Law for a system of particles dP Fnet dt Conservation of linear momentum • From the Newton’s Second Law dP Fnet dt • If the net force acting on a system is zero, then dP 0 dt P const • If no net external force acts on a system of particles, the total linear momentum of the system is conserved (constant) • This rule applies independently to all components Fnet , x 0 Px const Chapter 9 Problem 17 A popcorn kernel at rest in a hot pan bursts into two pieces, with masses 91 mg and 64 mg. The more massive piece moves horizontally at 47 cm/s. Describe the motion of the second piece. Impulse • During a collision, an object is acted upon by a force exerted on it by other objects participating in the collision dp Fnet (t ) dt dp Fnet (t )dt tf ti tf dp Fnet (t )dt • We define impulse as: tf J Fnet (t )dt ti • Then (momentum-impulse theorem) p f pi J ti Elastic and inelastic collisions • During a collision, the total linear momentum is always conserved if the system is isolated (no external force) • It may not necessarily apply to the total kinetic energy • If the total kinetic energy is conserved during the collision, then such a collision is called elastic • If the total kinetic energy is not conserved during the collision, then such a collision is called inelastic • If the total kinetic energy loss during the collision is a maximum (the objects stick together), then such a collision is called perfectly inelastic Elastic collision in 1D P const K const 2 2 m1v12i m2 v22i m1v1 f m2 v2 f 2 2 2 2 m1v1i m2 v2i m1v1 f m2v2 f m1 (v1i v1 f )(v1i v1 f ) m1 (v1i v1 f ) m2 (v2i v2 f ) m2 (v2i v2 f )(v2i v2 f ) (m1 m2 ) 2m2 v1 f v1i v2 i (m1 m2 ) (m1 m2 ) v2 f 2m1 (m2 m1 ) v1i v2 i (m1 m2 ) (m1 m2 ) Elastic collision in 1D: stationary target (m1 m2 ) 2m2 v1 f v1i v2 i (m1 m2 ) (m1 m2 ) 2m1 (m2 m1 ) v2 f v1i v2 i (m1 m2 ) (m1 m2 ) • Stationary target: v2i = 0 • Then (m1 m2 ) v1 f v1i (m1 m2 ) 2m1 v2 f v1i (m1 m2 ) Perfectly inelastic collision in 1D P const m1v1i m2 v2i (m1 m2 )v f m1v1i m2 v2i vf m1 m2 Collisions in 2D P const Px const Py const Chapter 9 Problem 86 In a ballistic pendulum demonstration gone bad, a 0.52-g pellet, fired horizontally with kinetic energy 3.25 J, passes straight through a 400-g Styrofoam pendulum block. If the pendulum rises a maximum height of 0.50 mm, how much kinetic energy did the pellet have after emerging from the Styrofoam? Questions? Answers to the even-numbered problems Chapter 9 Problem 12 2.5 m Answers to the even-numbered problems Chapter 9 Problem 16 4680 km Answers to the even-numbered problems Chapter 9 Problem 18 – 10.6 iˆ – 2.8 jˆ m/s Answers to the even-numbered problems Chapter 9 Problem 78 7.95 s